Question: The equation of a circle $C$ is $x^2+y^2+6x-14y+9 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2+6x) + (y^2-14y) = -9$ $(x^2+6x+9) + (y^2-14y+49) = -9 + 9 + 49$ $(x+3)^{2} + (y-7)^{2} = 49 = 7^2$ Thus, $(h, k) = (-3, 7)$ and $r = 7$.